The value of tan 81^{circ}-tan 63^{circ}-tan | vedclass

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Question

(d)

We have,

$	an 81^{\circ}-	an 63^{\circ}-	an 27^{\circ}+	an 9^{\circ}$

$=\left(	an 81^{\circ}+	an 9^{\circ}ight)-\left(	an 63^{\c

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d)

We have,

$	an 81^{\circ}-	an 63^{\circ}-	an 27^{\circ}+	an 9^{\circ}$

$=\left(	an 81^{\circ}+	an 9^{\circ}ight)-\left(	an 63^{\circ}+	an 27^{\circ}ight)$

$=\left(\cot 9^{\circ}+	an 9^{\circ}ight)-\left(\cot 27^{\circ}+	an 27^{\circ}ight)$

$=\left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}}+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}ight)-\left(\frac{\cos 27^{\circ}}{\sin 27^{\circ}}+\frac{\sin 27^{\circ}}{\cos 27^{\circ}}ight)$

$=\left(\frac{\cos ^2 9^{\circ}+\sin ^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}ight)-\left(\frac{\cos ^2 27^{\circ}+\sin ^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}ight)$

$=\left(\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}ight)-\left(\frac{2}{2 \sin 27^{\circ} \cos 27^{\circ}}ight)$

$=\left[\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}ight]$

$=2\left[\frac{1}{\sin 18^{\circ}}-\frac{1}{\cos 36^{\circ}}ight]=2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}ight]$

$\left[\because \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}, \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}ight]$

$=8\left[\frac{\sqrt{5}+1-\sqrt{5}+1]}{5-1}ight]=4$

The value of $\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$ is

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